where \(\hat{H^{\prime }}\)  includes excitation between \(\lbrace |1r\rangle ,~|r1\rangle \rbrace \)  and \(|rr\rangle \)  and the dipole-dipole flip from \(|rr\rangle \) . Focusing on the strong interaction regime, \(\hat{H^{\prime }}\)  can be discarded because \(|rr\rangle \)  is not coupled [1]} and hence \( \hat{H}=\bar{\Omega }|\mathbb {R}\rangle \langle 11|/2 +\text{H.c.}\) , where \(\bar{\Omega }\equiv \sqrt{\Omega _{\text{c}}^2+\Omega _{\text{t}}^2}\)  and \(|\mathbb {R}\rangle \equiv (\Omega _{\text{c}}|r1\rangle + \Omega _{\text{t}}|1r\rangle )/\bar{\Omega }\) . When \(\alpha \equiv \Omega _{\text{c}}/\Omega _{\text{t}}=\sqrt{15}\) , we have \(\bar{\Omega }t_1 = 4\pi \) , so that the wavefunction at \(t=t_0+t_1\) , given by \(e^{-it_1 \hat{H}}|\psi (t_0)\rangle \) , is equal to Eq. (REF ), and it is like that nothing happens to the target qubit upon the completion of the drive in the control qubit. Then, because the pulse for the control qubit ends at \(t=t_0+t_1\) , the continuous pumping on the target qubit drives the state to \(|1r\rangle \)  at \(t=2t_1\) , i.e., a \(2\pi \)  pulse, instead of a \(\pi \)  pulse, completes the transition \(|11\rangle \rightarrow |1r\rangle \) , which corresponds to a 2-fold slow-down.
<FIGURE><FIGURE>